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User blog:Katakana133/2 Notations that actually make sense now.
Okay, so that last one... In April... Yeah that wasn't very good. Today? I'll be making a notation and a slightly stronger notation. What are the two notations? These are versions of a similar set of rules for an array notation. Here are the basic rules for two series: The first one is one you may know: The Fibonacci sequence. The rule? F(0) = 1 and F(1) = 1. Iff n >= 2, F(n) = F(n-1) + F(n-2). This gives rise to the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144... Now, a different sequence! This one is simpler: S(0) = 0 and S(1) = 1. Iff n >= 2, S(n) = S(n-1) + n, otherwise known as the simple S(n) = 1 + 2 + 3 + 4 ... + (n - 1) + n. This gives rise to: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78... First, I'll extend the S(n) function. How? Use more dimensions! What do I mean? If you look up these numbers, you see that they're called triangular, or just triangle numbers, because it looks like they're arranged in a triangle. I add... The TETRAHEDRAL numbers!! What are they? Let's extend S(n) to S(n, m), where m is the dimension of the shape, minus 1. So S(n) = S(n, 1) and the tetrahedral numbers are S(n, 2). Now for the definition. S(n, m) = S(1, m-1) + S(2, m-1) + S(3, m-1) ... + S(n-1, m-1) + S(n, m-1). Now, another sequence! The sequence goes like this: S(1, 1), S(2, 2), S(3, 3), S(4, 4) and so on. This isn't an array notation, it's just some 2-argument pascal triangle calculator. Now, for the continuation of the Fibonacci sequence to the second dimension. Now, there are two rows. I'm removing the first 1 because that is F(0). I'm starting at F(1), which is important. So, here it is: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233... Now, for the second row. This is the algorithm. F(n, m) = F(F(n, m-1), m-1). This creates: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233... 1, 2, 3, 8, 34, 377... 1, 2, 3... ... This continues out to ... F(1, 1) F(2, 1) F(3, 1) F(4, 1) F(5, 1)... F(1, 2) F(2, 2) F(3, 2) F(4, 2) F(5, 2)... F(1, 3) F(2, 3) F(3, 3) F(4, 3) F(5, 3)... F(1, 4) F(2, 4) F(3, 4) F(4, 4) F(4, 5)... F(1, 5) F(2, 5) F(3, 5) F(4, 5) F(5, 5)... ... ... ... ... ... ... Okay, with that out of the way, this notation is at the tetrational level (If I'm calculating correctly). Now, the fω(n) level. This requires adding another argument to the F function. So, this fits nicely with the third dimension. Just: F(x, y) = F(x, y, 1), F(n, 1, m) = F(n, n, m-1). I'll have to make the rules less spread out so you can understand the notation a little bit more: TERMINOLOGY: ,# means all unspecified entries after. #, means all unspecified entries before. I will start with some and gradually add more. ...x of y... means that x is repeated y times, in entries. EXAMPLE: F(#, 16, ...16 of F(3, 4)..., 16,#) 1. F(0) = 1. F(1) = 1. Iff n >= 2, F(n) = F(n-1) + F(n-2) 2. F(n, m,#) = F(F(n, m-1,#), m-1,#) 3. Iff y > 1, F(...x of y... , z-1,#) = F(x, ...1 of y-1... , z,#) 4. The default number in an array is 1. The last entry in the array you can see has to be 2 or larger. Otherwise, remove the entry. These four rules add an indefinite number of dimensions to the Fibonacci sequence! Under the assumption that this directly relates to the FGH with each plane (third entry) adding one to the ordinal in the FGH, this notation maxes out at fωω(n). This may or may not be correct, but this is what I think, so I'll continue out from here in a bit.